poj 1008:Maya Calendar（模拟题，玛雅日历转换）

时间：2023-01-22 15:39:06
Maya Calendar
Time Limit: 1000MS   Memory Limit: 10000K
Total Submissions: 64795   Accepted: 19978

Description

During his last sabbatical, professor M. A. Ya made a surprising discovery about the old Maya calendar. From an old knotted message, professor discovered that the Maya civilization used a 365 day long year, called Haab, which had 19 months. Each of the first 18 months was 20 days long, and the names of the months were pop, no, zip, zotz, tzec, xul, yoxkin, mol, chen, yax, zac, ceh, mac, kankin, muan, pax, koyab, cumhu. Instead of having names, the days of the months were denoted by numbers starting from 0 to 19. The last month of Haab was called uayet and had 5 days denoted by numbers 0, 1, 2, 3, 4. The Maya believed that this month was unlucky, the court of justice was not in session, the trade stopped, people did not even sweep the floor.

For religious purposes, the Maya used another calendar in which the year was called Tzolkin (holly year). The year was divided into thirteen periods, each 20 days long. Each day was denoted by a pair consisting of a number and the name of the day. They used 20 names: imix, ik, akbal, kan, chicchan, cimi, manik, lamat, muluk, ok, chuen, eb, ben, ix, mem, cib, caban, eznab, canac, ahau and 13 numbers; both in cycles.

Notice that each day has an unambiguous description. For example, at the beginning of the year the days were described as follows:

1 imix, 2 ik, 3 akbal, 4 kan, 5 chicchan, 6 cimi, 7 manik, 8 lamat, 9 muluk, 10 ok, 11 chuen, 12 eb, 13 ben, 1 ix, 2 mem, 3 cib, 4 caban, 5 eznab, 6 canac, 7 ahau, and again in the next period 8 imix, 9 ik, 10 akbal . . .

Years (both Haab and Tzolkin) were denoted by numbers 0, 1, : : : , where the number 0 was the beginning of the world. Thus, the first day was:

Haab: 0. pop 0

Tzolkin: 1 imix 0 
Help professor M. A. Ya and write a program for him to convert the dates from the Haab calendar to the Tzolkin calendar. 

Input

The date in Haab is given in the following format: 
NumberOfTheDay. Month Year

The first line of the input file contains the number of the input dates in the file. The next n lines contain n dates in the Haab calendar format, each in separate line. The year is smaller then 5000.

Output

The date in Tzolkin should be in the following format: 
Number NameOfTheDay Year

The first line of the output file contains the number of the output dates. In the next n lines, there are dates in the Tzolkin calendar format, in the order corresponding to the input dates.

Sample Input

3

10. zac 0

0. pop 0

10. zac 1995

Sample Output

3

3 chuen 0

1 imix 0

9 cimi 2801

Source

 
　　水题，模拟题。
　　题意真让人无语，把我这个英语渣渣虐得死去活来……好吧，鄙视我吧
　　参考博客：POJ 1008 Maya Calendar     POJ1008-Maya Calendar
　　题意：

Habb历一年365天

Tzolkin历一年260天

先计算Habb历从第0天到输入日期的总天数sumday

Sumday/day就是Tzolkin历的年份

Tzolkin历的天数Name每20一循环，先建立Tzolkin历天数Name与1~20的映射，

因此Sumday %20+1就是Tzolkin历的天数Name

Tzolkin历的天数ID每13一循环，且从1开始，则Sumday %13+1就是Tzolkin历的天数ID

　　另外Habb历是一共有19个月，前18个月每月20天，最后一月只有5天，加起来一年一共365天；
　　Tzolkin历一共有13个月，每月20天，一年一共260天。尽管有20个名字，正好对应每月20天，但是Tzolkin历的天数编号有些坑。例如第一个月20天不是从1到20编号，而是从1到13再从1到7，即第一个月编号是这样的：1、2、3、4、5、6、7、8、9、10、11、12、13、1、2、3、4、5、6、7。下一个月再从8开始。
　　输入时Habb历的天数，月份名字，年份。
　　最后让你输出Tzolkin历的天数编号，对应的天数名字，年份。（不要求输出月份）
　　思路：
　　先根据输入的Habb历的天数，月份（需要将名字转换成数字），年份，求出从0开始度过的总天数sumday。
　　然后转换成Tzolkin历的天数编号dayId，天数名字dayName，年份year。
　　求法分别为：
　　year = sumday/;

　　dayName = sumday%;　　//这只是下标，最后输出对应的天数名字

　　dayId = sumday%+;

　　最后输出：

    cout<<dayId<<' '<<Tzolkin[dayName]<<' '<<year<<endl;

　　代码：

 #include <iostream>

 #include <string.h>

 using namespace std;

 char Habb[][] =

 {"pop", "no", "zip", "zotz", "tzec",

  "xul", "yoxkin", "mol", "chen", "yax",

  "zac", "ceh", "mac", "kankin", "muan",

  "pax", "koyab", "cumhu","uayet"};

 char Tzolkin[][] =

 {"imix", "ik", "akbal", "kan", "chicchan",

  "cimi", "manik", "lamat", "muluk", "ok",

  "chuen", "eb", "ben", "ix", "mem",

  "cib", "caban", "eznab", "canac", "ahau"

 };

 int Name2month(char name[]) //返回该名字对应的月份

 {

     for(int i=;i<;i++){  //依次比较

         if(strcmp(Habb[i],name)==)

             return i+;

     }

     return -;

 }

 int GetHabbSumday(int day,int month,int year)   //获得总天数

 {

     int sumday = ;

     sumday += year*;

     sumday += *(month-);

     sumday += day;

     return sumday;

 }

 int main()

 {

     int n,Number,Year;

     char c,Name[];

     cin>>n;

     cout<<n<<endl;

     while(n--){

         cin>>Number>>c>>Name>>Year; //输入

         int sumday = GetHabbSumday(Number,Name2month(Name),Year);   //获得总天数

         //计算结果

         int year = sumday/;

         int dayName = sumday%;

         int dayId = sumday%+;

         cout<<dayId<<' '<<Tzolkin[dayName]<<' '<<year<<endl;

     }

     return ;

 }

Freecode : www.cnblogs.com/yym2013

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